# Electric Circuits Nilsson 7th Solution Manual

**Topics:**Trigraph, Orders of magnitude, Gh

**Pages:**18 (4535 words)

**Published:**May 6, 2013

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Assessment Problems

AP 1.1 To solve this problem we use a product of ratios to change units from dollars/year to dollars/millisecond. We begin by expressing $10 billion in scientiﬁc notation: $100 billion = $100 × 109 Now we determine the number of milliseconds in one year, again using a product of ratios: 1 year 1 hour 1 min 1 sec 1 year 1 day · · · = · 31.5576 × 109 ms 365.25 days 24 hours 60 mins 60 secs 1000 ms Now we can convert from dollars/year to dollars/millisecond, again with a product of ratios: 1 year 100 $100 × 109 · = = $3.17/ms 1 year 31.5576 × 109 ms 31.5576 AP 1.2 First, we recognize that 1 ns = 10−9 s. The question then asks how far a signal will travel in 10−9 s if it is traveling at 80% of the speed of light. Remember that the speed of light c = 3 × 108 m/s. Therefore, 80% of c is (0.8)(3 × 108 ) = 2.4 × 108 m/s. Now, we use a product of ratios to convert from meters/second to inches/nanosecond: 100 cm 1 in (2.4 × 108 )(100) 9.45 in 2.4 × 108 m 1s · · = = · 9 1s 10 ns 1m 2.54 cm (109 )(2.54) 1 ns Thus, a signal traveling at 80% of the speed of light will travel 9.45 in a nanosecond.

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1–2 AP 1.3

CHAPTER 1. Circuit Variables Remember from Eq. (1.2), current is the time rate of change of charge, or i = dq In dt this problem, we are given the current and asked to ﬁnd the total charge. To do this, we must integrate Eq. (1.2) to ﬁnd an expression for charge in terms of current: q(t) = t 0

i(x) dx

We are given the expression for current, i, which can be substituted into the above expression. To ﬁnd the total charge, we let t → ∞ in the integral. Thus we have qtotal = = AP 1.4 ∞ 0

20e−5000x dx =

20 −5000x e −5000

∞ 0

=

20 (e∞ − e0 ) −5000

20 20 (0 − 1) = = 0.004 C = 4000 µC −5000 5000

Recall from Eq. (1.2) that current is the time rate of change of charge, or i = dq . In dt this problem we are given an expression for the charge, and asked to ﬁnd the maximum current. First we will ﬁnd an expression for the current using Eq. (1.2): i= = dq d 1 1 t = + 2 e−αt − 2 dt dt α α α d 1 d t −αt d 1 −αt e − − e dt α2 dt α dt α2 1 −αt t 1 − α e−αt − −α 2 e−αt e α α α 1 1 −αt e +t+ α α

= 0− = −

= te−αt Now that we have an expression for the current, we can ﬁnd the maximum value of the current by setting the ﬁrst derivative of the current to zero and solving for t: d di = (te−αt ) = e−αt + t(−α)eαt = (1 − αt)e−αt = 0 dt dt Since e−αt never equals 0 for a ﬁnite value of t, the expression equals 0 only when (1 − αt) = 0. Thus, t = 1/α will cause the current to be maximum. For this value of t, the current is i= 1 −α/α 1 = e−1 e α α 1 e−1 ∼ 10 A = 0.03679

Remember in the problem statement, α = 0.03679. Using this value for α, i=

Problems AP 1.5 Start by drawing a picture of the circuit described in the problem statement:

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Also sketch the four ﬁgures from Fig. 1.6:

[a] Now we have to match the voltage and current shown in the ﬁrst ﬁgure with the polarities shown in Fig. 1.6. Remember that 4A of current entering Terminal 2 is the same as 4A of current leaving Terminal 1. We get (a) v = −20 V, (c) v = 20 V, i = −4 A; i = −4 A; (b) v = −20 V, (d) v = 20 V, i = 4A i = 4A

[b] Using the reference system in Fig. 1.6(a) and the passive sign convention, p = vi = (−20)(−4) = 80 W. Since the power is greater than 0, the box is absorbing power. [c] From the calculation in part (b), the box is absorbing 80 W. AP 1.6 Applying the passive sign convention to the power equation using the voltage and current polarities shown in Fig. 1.5, p = vi. From Eq. (1.3), we know that power is the time rate of change of energy, or p = dw . If we know the power, we can ﬁnd the dt energy by integrating Eq. (1.3). To begin, ﬁnd the expression for power: p = vi = (10,000e−5000t )(20e−5000t ) = 200,000e−10,000t = 2 × 105 e−10,000t W Now ﬁnd the expression for energy by integrating Eq. (1.3): w(t) = t 0

p(x) dx

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CHAPTER 1. Circuit Variables Substitute the...

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